Question

From the following equation, find the value of constant 'k' so that the equation has real and equal roots.

(k+ 4)x² + (k + 1)x + 1 = 0

Answer 1

Given,

(k + 4)x² + (k + 1)x + 1 = 0

Comparing (k + 4)x² + (k + 1)x + 1 = 0 with ax² + bx + c = 0 we get,

a = k + 4, b = k + 1 and c = 1.

Since roots are real and equal,

We know that,

∴ D = 0

⇒ b² − 4ac = 0

⇒ (k + 1)² − 4(k + 4)(1) = 0

⇒ k² + 2k + 1 − 4k − 16 = 0

⇒ k² − 2k − 15 = 0

⇒ k² − 5k + 3k − 15 = 0

⇒ k(k − 5) + 3(k − 5) = 0

⇒ (k + 3)(k − 5) = 0

⇒ k + 3 = 0 or k − 5 = 0

⇒ k = −3 or k = 5.

If k = 5, substituting in (k + 4)x² + (k + 1)x + 1 = 0 we get,

⇒ (5 + 4)x² + (5 + 1)x + 1 = 0

⇒ 9x² + 6x + 1 = 0

⇒ (3x + 1)² = 0

⇒ 3x + 1 = 0

⇒ x = `-1/3`

Both roots are the same.

If k = −3, substituting in (k + 4)x² + (k + 1)x + 1 = 0 we get,

⇒ (−3 + 4)x² + (-3 + 1)x + 1 = 0

⇒ x² − 2x + 1 = 0

⇒ (x − 1)² = 0

⇒ x − 1 = 0

⇒ x = 1

Both roots are the same.

Hence, k = −3 or 5